# Two satellites P_"1" and P_"2" are revolving in orbits of radii R and 4R. The ratio of maximum and minimum angular velocities of the line joining P_"1" and P_"2" is ??

Jun 16, 2018

$- \frac{9}{5}$

#### Explanation:

According to Kepler's third law, ${T}^{2} \propto {R}^{3} \implies \omega \propto {R}^{- \frac{3}{2}}$, if the angular velocity of the outer satellite is $\omega$, that of the inner one is $\omega \times {\left(\frac{1}{4}\right)}^{- \frac{3}{2}} = 8 \omega$.

Let us consider $t = 0$ to be an instant when the two satellites are collinear with the mother planet, and let us take this common line as the $X$ axis. Then, the coordinates of the two planets at time $t$ are $\left(R \cos \left(8 \omega t\right) , R \sin \left(8 \omega t\right)\right)$ and $\left(4 R \cos \left(\omega t\right) , 4 R \sin \left(\omega t\right)\right)$, respectively.

Let $\theta$ be the angle the line joining the two satellites makes with the $X$ axis. It is easy to see that

$\tan \theta = \frac{4 R \sin \left(\omega t\right) - R \sin \left(8 \omega t\right)}{4 R \cos \left(\omega t\right) - R \cos \left(8 \omega t\right)} = \frac{4 \sin \left(\omega t\right) - \sin \left(8 \omega t\right)}{4 \cos \left(\omega t\right) - \cos \left(8 \omega t\right)}$

Differentiation yields

${\sec}^{2} \theta \frac{d \theta}{\mathrm{dt}} = \frac{d}{\mathrm{dt}} \frac{4 \sin \left(\omega t\right) - \sin \left(8 \omega t\right)}{4 \cos \left(\omega t\right) - \cos \left(8 \omega t\right)}$

$= {\left(4 \cos \left(\omega t\right) - \cos \left(8 \omega t\right)\right)}^{-} 2 \times$
qquad [(4 cos(omega t)-cos(8 omega t))(4 omega cos(omega t)-8omega cos(8 omega t))-
qquad (4 sin(omega t)-sin(8 omega t))(-4omega sin(omega t)+8 omega sin(8 omega t)) ]

Thus

${\left(4 \cos \left(\omega t\right) - \cos \left(8 \omega t\right)\right)}^{2} \left[1 + {\left(\frac{4 \sin \left(\omega t\right) - \sin \left(8 \omega t\right)}{4 \cos \left(\omega t\right) - \cos \left(8 \omega t\right)}\right)}^{2}\right] \frac{d \theta}{\mathrm{dt}}$
 = 4 omega [(4 cos^2(omega t)-9 cos(omega t)cos (8 omega t) + 2 cos^2 (omega t))
qquad qquad + (4 sin^2(omega t)-9 sin(omega t)cos (8 omega t) +2sin^2 (omega t))]
$= 4 \omega \left[6 - 9 \cos \left(7 \omega t\right)\right] \implies$

$\left(17 - 8 \cos \left(7 \omega t\right)\right) \frac{d \theta}{\mathrm{dt}} = 12 \omega \left(2 - 3 \cos \left(7 \omega t\right)\right) \implies$

$\frac{d \theta}{\mathrm{dt}} = 12 \omega \frac{2 - 3 \cos \left(7 \omega t\right)}{17 - 8 \cos \left(7 \omega t\right)} \equiv 12 \omega f \left(\cos \left(7 \omega t\right)\right)$

Where the function

$f \left(x\right) = \frac{2 - 3 x}{17 - 8 x} = \frac{3}{8} - \frac{35}{8} \frac{1}{17 - 8 x}$

has the derivative

${f}^{'} \left(x\right) = - \frac{35}{17 - 8 x} ^ 2 < 0$

and is hence monotonically decreasing in the interval $\left[- 1 , 1\right]$ .

Thus, the angular velocity $\frac{d \theta}{\mathrm{dt}}$ is maximum when $\cos \left(7 \omega t\right)$ is minimum, and vice versa.

So,

${\left(\frac{d \theta}{\mathrm{dt}}\right)}_{\text{max}} = 12 \omega \frac{2 - 3 \times \left(- 1\right)}{17 - 8 \times \left(- 1\right)}$
$q \quad q \quad q \quad q \quad = 12 \omega \times \frac{5}{25} = \frac{12}{5} \omega$

${\left(\frac{d \theta}{\mathrm{dt}}\right)}_{\text{min}} = 12 \omega \frac{2 - 3 \times 1}{17 - 8 \times 1}$
$q \quad q \quad q \quad q \quad = 12 \omega \times \frac{- 1}{9} = - \frac{4}{3} \omega$

and so the ratio between the two is :

$\frac{12}{5} \omega : - \frac{4}{3} \omega = - 9 : 5$

Note The fact that $\frac{d \theta}{\mathrm{dt}}$ changes sign is the cause for so called apparent retrograde motion